I am told that the probability of measuring $\lambda$ is $$p_\lambda = Tr(\hat{P}_\lambda\hat{\rho}) = Tr(\hat{P}_\lambda\hat{\rho}\hat{P}_\lambda)$$ where $\hat{P}_\lambda = \sum_{n:\lambda_n = \lambda}|n\rangle\langle n|$ is the projection operator for eigenstates $n$ with an eigenvalue $\lambda$.
I have no idea how this is derived or why $$Tr(\hat{P}_\lambda\hat{\rho}) = Tr(\hat{P}_\lambda\hat{\rho}\hat{P}_\lambda).$$
This is not derived, this is the Born rule or the Lüders-von Neumann measurement rule. Either way, it is a fundamental axiom of quantum mechanics that measurements work this way. See e.g. this, this or this question for more discussion of possible motivations.
If you are fine with the Born rule for pure states and wonder where this rule for mixed states comes from, write the mixed state as a mixture of pure states $\rho = \sum_i p_i \lvert \psi_i\rangle\langle \psi_i\rvert$ and consider that the probability to measure the eigenvalue $\lambda$ for some observable $\Lambda$ with eigenstates $\lvert \lambda\rangle$ in the state $\lvert \psi_i\rangle$ would be $\lvert \langle \lambda\vert \psi_i\rangle \rvert^2$ by the Born rule and $$ \mathrm{tr}(\lvert \lambda\rangle \langle \lambda\vert \psi_i\rangle\langle \psi_i\rvert) = \lvert\langle \lambda\vert \psi_i \rangle\rvert^2,$$ so $$\mathrm{tr}(P_\lambda \rho) = \sum_i p_i \lvert\langle \lambda\vert \psi_i \rangle\rvert^2,$$ is just the probability to measure $\lambda$ in each of the $\psi_i$, weighted by their mixed probability $p_i$.
$\mathrm{tr}(P\rho P) = \mathrm{tr}(P^2\rho) = \mathrm{tr}(P\rho)$, where the first equality is due to the cyclicity of the trace ($\mathrm{tr}(ABC) = \mathrm{tr}(CAB)$ for any operators $A,B,C$) and the second because projections are idempotent ($P^2 = P$).
